Both products in Eq. (REF ) can be evaluated in a straigtforward manner.[1]} On one hand:
\(\begin{split}\log \triangle &=\sum \limits _{n=-\infty }^{+\infty }\frac{-i k^2}{4(\alpha n^2-\beta )}=\frac{-i k^2}{4\alpha }\sum \limits _{n=-\infty }^{+\infty }\frac{1}{n^2-\frac{\beta }{\alpha }}\\&=\frac{i k^2 \pi }{4\sqrt{\alpha \beta }} \cot {\pi \sqrt{\frac{\beta }{\alpha }}}\end{split}\) 
